Saved Bookmarks
| 1. |
Consider a quadraticequaiton az^(2) + bz + c=0, where a,b,c arecomplex number. The condition that theequation has onepurely imaginary root is |
|
Answer» `(cbara -abarc)^(2)= (b barc+ cbarb)(abara - bara b)` `z_(1) = - barz_(1)` and`underline(az_(1)^(2) + bz_(1) + c)=0""(1)` `rArr az_(1)^(2) + bz_(1) + c = 0` `rArr bara barz_(1)^(2) + barb barz_(1) + c = 0` `rArr bar z bar z_(1)^(2) + bar b bar z_(1) + barc = 0` `rArr bar a bar z_(1)^(2) - bar b barz_(1) + barc = 0""(as barz_(1) = - z_(1))""(2)` Now Eqs. (1) and (2) musthave one common root. `therefore ( cbara-abarc)^(2) =(barbc+ cbarb) (-abarb - barab)` Let `z_(1)` and `z_(2)` be two purely IMAGINARY ROOTS. Then, `barz_(1) = -z_(1), barz_(2) = - z_(2)` Now , `underline(abarz^(2) + bz + c) = 0""(3)` or `AZ^(2) + bz + c=bar0` or `bara barz_(20 + barb barz + barc =0` or `bara z^(2) - barbz + barc = 0""(4)` Equations (3) and (4) must be identical as their roots are same. ` therefore (a)/(bara) = -(b)/(barb)=(c)/(barc)` `rArr abarc = barac,+ barab = 0` and `b barc +barbc=0` . Hence, `barac` is purely real and `abarb` and `bbarc`are purely imaginary . let `z_(1)` (purely real ) be a root of the givenequation . Then , `z_(1) = barz_(1)`LTBR gt and ``underline(az_(1)^(2) + bz_(1) + c)= bar0""(5)` or `az_(1)^(2) + bz_(1) + c=0` or `baraz_(1)^(2) + bz_(1) + c = bar0` or `baraz_(1)^(2) + barb z_(1) + c= 0""(6)` Now(5) and (6) must have one common root. Hence, `(cbara - abarc)^(2) = (b barc - cbarb)(abarb-barab)` |
|